I know precisely what you mean but I'm afraid I don't know how to answer you! What you're saying makes all the sense but I was following that link! I'll talk to my teacher later today...

But in the "link" they do not have the load resistance on the diagram.

And once more, voltage drop across R_load is equal to voltage drop across Rc because they're also at a parallel setup, right?

Yes, exactly. But the voltage gain will drop because now Ic have to split between Rc and RL.
Also for Ai you have to include RL and Rth=(Rb1||Rb2) (Rbb in the link).

 

Ok, I'm going to check that later with my tutor. I'll let you know later today after classes or tomorrow when I get here!

 

Ok, after talking to my teacher I need to recalculate aV, aI, rIN and rOUT.

So far I have done the following:

hie = 25mV/Ib = 25/(4.187/458.7) ≈ 2.7kΩ

aV = (vOUT)/(vIN) = -(hfe*Ib*(Rc//Rl))/(hie*Ib) = (-hfe*(Rc//Rl))/hie = -458.7*1.67kΩ/2.7kΩ=-283.71

Now I'm not sure about current gain. I'm trying to analyse the circuit and I'm getting the following:

aI = iOUT/iIN = -((Rc*hfe*Ib)/(Rc+Rl))/Ib = -(Rc*hfe)/(Rc+Rl)

For iOUT I used a current divider for Rl, so would be (Rc*Ic)/Rc+Rl with Ic = hfe*Ib => (Rc*hfe*Ib)/(Rc+Rl).

Is the current gain correct?

 

hie = 25mV/Ib = 25/(4.187/458.7) ≈ 2.7kΩ
aV = (vOUT)/(vIN) = -(hfe*Ib*(Rc//Rl))/(hie*Ib) = (-hfe*(Rc//Rl))/hie = -458.7*1.67kΩ/2.7kΩ=-283.71

Are you sure ?? If RL = 2.2kΩ and Rc = 2.2kΩ then Rc||RL for sure cannot be equal to 1.67kΩ

Now I'm not sure about current gain. I'm trying to analyse the circuit and I'm getting the following:

aI = iOUT/iIN = -((Rc*hfe*Ib)/(Rc+Rl))/Ib = -(Rc*hfe)/(Rc+Rl)

For iOUT I used a current divider for Rl, so would be (Rc*Ic)/Rc+Rl with Ic = hfe*Ib => (Rc*hfe*Ib)/(Rc+Rl).

Is the current gain correct?

Looks good but what about Rb1 and Rb2 ?? Input current will also split between Rb1||Rb2 and hie.

 

Are you sure ?? If RL = 2.2kΩ and Rc = 2.2kΩ then Rc||RL for sure cannot be equal to 1.67kΩ

You right... It is 1.10kΩ. I'll change it when I get home.

Looks good but what about Rb1 and Rb2 ?? Input current will also split between Rb1||Rb2 and hie.

Well, I've been discussing that with some other mates in IRC (the old IRC yes). They are trying to help me but to be honest, I still haven't got anywhere about iIN.

I was thinking about writing iIN = vIN/Rin but that will lead me into an huge and ugly expression.

vIN = hfe*Ib

Rin = Rb1//Rb2//hie

so iIN = (hfe*Ib)/(Rb1//Rb2//hie)
and iOUT = (Rc*hfe*Ib)/(Rc + Rload)

But if I divide both terms as iOUT/iIN I'll get an expression with no current in it! Is it OK?

This is what I get after simplification:

Ai = iOUT/iIN = iOUT/(vIN/rIN) = ((Rc*hie*Ib)/(Rc + Rload))/((hie*Ib)/(Rb1//Rb2//hie))
= (Rc(Rb1//Rb2//hie))/(Rc + Rload)


Edited;

My teacher said that iIN is not Ib... Its Ib + the current across Rb1//Rb2.

 

Take a look here
http://www.zen22142.zen.co.uk/Theory/tr_model.htm (start from Current Gain Ai )

 

Take a look here
http://www.zen22142.zen.co.uk/Theory/tr_model.htm (start from Current Gain Ai )

I have already read that but there is no Rload in that circuit!
I also don't understand why do they move iIN to under Ib in the expression.

 

LOL. But now we do not talk about RL.
Look at the diagram and notice how Iin is divided into Ib and Rbi and Rb2 current. And how this influence the current gain.

 

LOL. But now we do not talk about RL.
Look at the diagram and notice how Iin is divided into Ib and Rbi and Rb2 current. And how this influence the current gain.

I think I understand that. Ib will be related with iIN by Rbb/(Rbb+hie).

But I want to write iIN and not Ib/iIN.

And I tried two ways:

1 - iOUT /iIN and then try to write iIN by means of Ib and hie and Rb1//Rb2

2 - iOUT/iIN but write iIN by means of vIN/rIN

 

Simply assume that you connect a Iin = 1A current source at the input. What will be the Iload current in terms of a Iin.

 

Simply assume that you connect a Iin = 1A current source at the input. What will be the Iload current in terms of a Iin.

I don't know what you mean, or better, I don't know if I understood your question. But that 1A of iIN will slpit into (Rb1//Rb2) and into hie. Then, the fraction that goes across hie is multiplied by hfe and lastly this current, so called Ic, will go across Rload, I guess.

 

But that 1A of iIN will slpit into (Rb1//Rb2) and into hie. Then, the fraction that goes across hie is multiplied by hfe and lastly this current, so called Ic, will go across Rload, I guess.

Very good. But notice that Ic also split into and Rc and into Rload.
And now you can use a current divider rule and find Iload in terms of this 1A input current.

 

Very good. But notice that Ic also split into and Rc and into Rload.
And now you can use a current divider rule and find Iload in terms of this 1A input current.

Ok, but I'm deviating from what I really need. I'm sorry but I have to try to write the formula for that circuit.

Check if I can do the following:

 

Looks good. But did you forget about "minus" sing ??

But notice that if you assume Rbb= Rb1||Rb2 and use a current divider rule we can write

Iout = Iin * Rbb/(Rbb + Hie) * Hfe * -Rc/(Rc + RL) and directly from this we have

Iout/Iin = - (Hfe Rbb Rc)/((Hie + Rbb) * (Rc + RL))

 

Looks good. But did you forget about "minus" sing ??

But notice that if you assume Rbb= Rb1||Rb2 and use a current divider rule we can write

Iout = Iin * Rbb/(Rbb + Hie) * Hfe * -Rc/(Rc + RL) and directly from this we have

Iout/Iin = - (Hfe Rbb Rc)/((Hie + Rbb) * (Rc + RL))

I didn't understood the expression for Iout after Hfe... I understand that "Iin * Rbb/(Rbb + Hie) * Hfe" but then you have a * and a - signs together without any curly braces... Is anything missing there?

And is your final expression for Iout/Iin equal to mine? My teacher told me that current gain is positive and voltage gain is negative!


Edited;
Ok, I think I understood your equations but I think they are not equal to what I got!

Edited2;
Ok, result is the same!

 

I didn't understood the expression for Iout after Hfe... I understand that "Iin * Rbb/(Rbb + Hie) * Hfe" but then you have a * and a - signs together without any curly braces... Is anything missing there?

And is your final expression for Iout/Iin equal to mine? My teacher told me that current gain is positive and voltage gain is negative!


Edited;
Ok, I think I understood your equations but I think they are not equal to what I got!

Edited2;
Ok, result is the same!

It's possible to derive a fairly simple expression for a current divider when more than two resistors are involved. For example, suppose we have 3 resistors in parallel, R1, R2 and R3. We apply a voltage V to the three and we want to know what fraction of the total current flows in R2. We can proceed like this, and it's easy to see what things look like if we have "n" resistors instead of just three:

The solution for the current gain Ai can take several forms, but all of them are equivalent to multiplying the input current by hfe and then by a current divider at the input, followed by a current divider at the output.

PsyScOrpiOn's instructor apparently has used a different definition for the sign of the output current.

Here I compare my solution, PsyScOrpiOn's solution and Jony130's solution. For all of them, I've shown the input current divider part in red and the output current divider part in blue. I then simplified all of them and show that the final result is the same for all with the substitution Rbb = Rb1||Rb2.

 

Thanks for the thoroughly explanation. I replaced the letters by their values and checked that my equation and Jony130's were returning the same result!

Now I need to analyse the circuit without bypassing CE so that I can find AC Load Line.

It was easy for the previous circuit where Re was out of the circuit because of CE but now we have to consider it.
My problem is that I have more unknows now and I don't know how to get rid of them! I'm talking about ie (AC value) and IE (DC value).

 

We have defined the following:

VCE - all capital letters to refer to DC value of voltage drop at C-E junction
vce - all regular letters to refer to AC value of voltage drop at C-E junction
vCE - capital index letters to refer to AC + DC values of voltage drop at C-E junction

The same for colector current and emitter current.

IC - DC value
ic - AC value
iC - AC + DC value

IE - DC value
ie - AC value
iE - AC + DC value

So, for the output net equation (AC analysis) I have:

vce + ic*(Rc//Rload) + Re*ie = 0

using the following relations:

iC = IC + ic <=> ic = iC-IC
vCE = VCE + vce <=> vce = vCE-VCE

and replacing them in the output net equation, I have

(vCE-VCE) + (iC-IC)*(Rc//Rload) + Re*(iE-IE) = 0

My question is how do I find iE. IE I know that is IC*(β+1)/β but I don't know if I can do the same for iE and say that iE = iC*(β+1)/β

 

OMG.
Ie = Ib + Ic = Ic*(β+1)/β no matter what.
Also notice that β is a current gain for DC and hfe is a current gain for AC signal.

 

OMG.
Ie = Ib + Ic = Ic*(β+1)/β no matter what.
Also notice that β is a current gain for DC and hfe is a current gain for AC signal.

Ok, I didn't know those two details.